- #1
nonequilibrium
- 1,439
- 2
Hello,
Differentiability of [itex]f : \mathbb C \to \mathbb C[/itex] is characterized as [itex]\frac{\partial f}{\partial z^*} = 0[/itex].
More exactly: [itex]\frac{\partial f(z,z^*)}{\partial z^*} := \frac{\partial f(z[x(z,z^*),y(z,z^*)])}{\partial z^*} = 0[/itex] where [itex]z(x,y) = x+iy[/itex] and [itex]x(z,z^*) = \frac{z+z^*}{2}[/itex] and analogously for y.
But anyway, does this make sense? More specifically, is it consistent? The thing I'm having trouble with is that it looks like that we can always make the differentation w.r.t. z* zero. For example, for clarity of my argument, define
[itex]g : \mathbb C \to \mathbb C: z \mapsto z^*[/itex]
then for the modulus function [itex] \frac{\partial |z|}{\partial z^*} = \frac{\partial \sqrt{z g(z)}}{\partial z^*} = 0[/itex] as g is a function of z and not z*.
Where is the mathematical error in this reasoning?
Differentiability of [itex]f : \mathbb C \to \mathbb C[/itex] is characterized as [itex]\frac{\partial f}{\partial z^*} = 0[/itex].
More exactly: [itex]\frac{\partial f(z,z^*)}{\partial z^*} := \frac{\partial f(z[x(z,z^*),y(z,z^*)])}{\partial z^*} = 0[/itex] where [itex]z(x,y) = x+iy[/itex] and [itex]x(z,z^*) = \frac{z+z^*}{2}[/itex] and analogously for y.
But anyway, does this make sense? More specifically, is it consistent? The thing I'm having trouble with is that it looks like that we can always make the differentation w.r.t. z* zero. For example, for clarity of my argument, define
[itex]g : \mathbb C \to \mathbb C: z \mapsto z^*[/itex]
then for the modulus function [itex] \frac{\partial |z|}{\partial z^*} = \frac{\partial \sqrt{z g(z)}}{\partial z^*} = 0[/itex] as g is a function of z and not z*.
Where is the mathematical error in this reasoning?